# Trigonometric inequalities

junaidnawaz

$$|\gamma| \leq \cos (\beta)$$
$$\arccos (|\gamma|) \leq \beta$$

does taking the arccos() on both sides of equation changes the relational operator??

Homework Helper
welcome to pf!

hi junaidnawaz! welcome to pf!

(have a beta: β and a gamma: γ and a ≤ )

arccos is defined as being in [0,π)

so long as β is also in [0,π), your equations are the same (because cos is monotone in that region, and therefore so is arccos)

junaidnawaz

in my case, the range of parameters is as,
$$0 \leq \beta \leq \pi /2$$
$$-1 \leq \gamma \leq +1$$

by taking arccos() on both-sides, would it change the operator (from $$\leq$$ to $$\geq$$ ) or would it remain same ??

Homework Helper
by taking arccos() on both-sides, would it change the operator (from $$\leq$$ to $$\geq$$ ) or would it remain same ??

oh, i missed that!

yes, cos is decreasing, so the ≤ changes to ≥

(but, eg, sin is increasing, so the ≤ would stay the same )

junaidnawaz
Thank you.

if
$$|\gamma| \leq \cos( \beta )$$

then

$$x \leq \beta$$

can i find "x", by keeping the RHS fixed to $$\beta$$

is this possible to find x ?? by keeping RHS and relation operator the same ??

Homework Helper
sorry, i don't understand …

isn't x just |γ| ?

junaidnawaz
$$|\gamma| \leq \cos ( \beta )$$

When I take arccos() on both sides, it becomes

$$\arccos( |\gamma| ) \geq \beta$$

however, i want to keep $$\beta$$ on right side, and i want to keep the relational operator as $$\leq$$, i.e.,

$$x \leq \beta$$

what would be x ??

junaidnawaz
I wounder if its not a stupid question .... :P

Homework Helper
however, i want to keep $$\beta$$ on right side, and i want to keep the relational operator as $$\leq$$, i.e.,

$$x \leq \beta$$

that's not possible (unless you replace β by some decreasing function of β, such as 1/β or -β)

junaidnawaz
Thank you :)